Defining an Ellipse
Date: 07/16/2003 at 15:51:18 From: Chris Subject: Defining an ellipse A right cylinder of radius "r" (a lift duct) intercepts a plane (the deck of a hovercraft) at an angle of 30 degrees to the vertical. How may the resulting ellipse be scribed? - i.e., how do I mark the hole to be cut in the deck?
Date: 07/16/2003 at 17:06:39 From: Doctor Peterson Subject: Re: Defining an ellipse Hi, Chris. There are a couple of steps to the question: first, we have to find the dimensions of the ellipse, and then we have to draw it. Here is a side view of your cylinder and plane; as I understand it, the deck is horizontal and the cylinder is tilted at a 30 degree angle to it: -+ ---- \ 2r --- \ --- \ ---- \ --- A \ ---+-------------------------------- |\ 2a \ | \ \ | \ \ | A \ \ | \ \ | \ \ | \ \ \ \ \ \ \ -+ \ ---- \ --- \ --- \ ---- \ --- +- Here A is the angle (30 degrees), r is the radius of the cylinder (so that 2r is its diameter), and 2a is the long axis of the ellipse. Note that the short axis of the ellipse will be 2r. Using simple trigonometry, 2r -- = cos(A) 2a so a = r/cos(A) We call a the semi-major axis of the ellipse; the semi-minor axis is called b, so b = r Now, how do we draw the ellipse? There are two best-known methods. One uses a pair of circles (with radii a and b) to plot points on the ellipse, and is described here: Accurate Drawing of an Ellipse http://mathforum.org/library/drmath/view/55085.html A quicker, but somewhat less accurate method, is given here: Ellipse Area and Circumference http://mathforum.org/library/drmath/view/55304.html and shown nicely here: Draw an ellipse, using pencil and string http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/conics/ drawing/ellipse_string.html The length of the string is our 2a, and the two nails or tacks (the foci) are a distance of 2c apart, where c^2 = a^2 - b^2 This fact may be seen here: | oooooBooooo----------------- oooo /|\ oooo ^ ooo / | \ ooo | oo / | \ oo | o / | \a oo b oo / b| \ o | o / | \ o | o / | \ o v ----o---------F-------O-------G---------A---- o | c o o | o| oo | o| o | oo | oo | oo | ooo | ooo | oooo | oooo | ooooooooooo | | | |<-------a------->| | | Since both the top point B and the side point A are drawn with the same total length of string, the lengths FB+GB = 2GB and FA+GA = 2OA are equal to 2a. This gives a right triangle GOB to which we can apply the Pythagorean theorem to get the fact I am demonstrating. So you can measure the string and the nails carefully (which is tricky because of the stretchiness of the string), or use the two- circle method to make a more precise, but more tedious, outline of the hole you need. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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